Identifying special solutions to ODEs

Question

I came across this ODE:

$ y' = 2(\frac{y}{x}) + 1 $

I managed to solve it absolutely fine, recognising it was a Homogenous first-order ODE, using the substitution $v = y/x$ etc.

My answer was -

$ y = Ax^2 - x $
where $A = \pm e^c $

See, that was where my set of solutions thwarted, however $ y = -x $ is considered a 'special solution'.

How would I go about identifying this, because $ \pm e^c \ne 0$ so I am struggling to figure it out on my own.

Any help is appreciated :)

Answer 1

When you are solving the equation you need to be careful of situations like $\frac{1}{v+1}$. What happens when $y = -x$?

Answer 2

$$y'-\frac{2}{x}y=1\implies\mu=e^{-2\operatorname{ln}x+C}=C_1x^{-2}$$

$$(y\mu)'=C_1x^{-2}\implies y=\frac{1}{\mu}(-C_1x^{-1}+C_2)=-x+C_3x^2$$

$C_3=\frac{C_2}{C_1}$

$C_2 \in \mathbb{R}$

$C_1=e^C$

$C\in\mathbb{R}$

$C_1\neq0$ but $C_2$ is any number so $C_3$ can be $0$